Problem Set

1 ∀x∀y∀z[(Ox ∨ ¬N y)&(Ox ∨ M z)] 2 ∀x∀y(Ox → ¬¬Ly) 3 ∀x∀y∀z[¬N x → (M y → Lz)] 4 (Oa ∨ ¬N a)&(Oa ∨ M a) ∀E, 1 5 Oa → ¬¬La ∀E, 2 6 ¬N a → (M a → La) ∀E, 3 7 ¬La Assume for RAA 8 ¬Oa MT, 5, 7 9 Oa ∨ ¬N a ∧E, 4 10 ¬N a ∨E, 8, 9 11 M a → La ⇒E, 6, 10 12 Oa ∨ M a ∧E, 4 13 Ma ∨E, 8, 12 14 La ⇒E, 11, 13 15 ¬La R, 7 16 La RAA, 7–15 17 ∃xLx ∃I, 16 1 Problem Set #4 1 Translations Translate each of the following sentences into QL, providing a symbolization key and specifying the UD when necessary. (5 pts. each) 1. Not all murderers deserve capital punishment, but Smith is a murderer who deserves capital punishment. 2. If Darwin is not a biologist, then no one is a biologist. 3. Something is red all over and something is green all over, but it is not true that something is both red and green all over. 4. Everyone is better than someone but no one is better than everyone. 5. Only those who love themselves are truly happy. 2 Proofs 6. Translate the following argument into QL, providing a symbolization key and specifying the UD when necessary, then construct a proof to show that the argument is valid. (10 pts. for translation, 15 pts. for proof) Carl is not on the team. For Carl is a sprinter, and Carl is faster than any sprinter on the team. But no one is faster than himself. 1 Problem Set #4 Answer Key 1. (a) M x = x is a murderer; Dx = x deserves capital punishment; s = Smith (b) ¬∀x(M x → Dx)&M s&Ds 2. (a) Bx = x is a biologist; d = Darwin (b) ¬Bd → ¬∃xBx 3. (a) Rx = x is red all over; Gx = x is green all over (b) ∃xRx&∃xGx&¬∃x(Rx&Gx) 4. (a) Bxy = x is better than y (b) ∀x∃yBxy&¬∃x∀yBxy 5. (a) Lxy = x loves y; Hx = x is truly happy (b) ∀x(Hx → Lxx) 6. T x = x is on the team; Sx = x is a sprinter; F xy = x is faster than y; c = Carl 1 Sc&∀x((Sx&T x) → F cx) 2 ∀x¬F xx 3 ∀x((Sx&T x) → F cx) ∧E, 1 4 (Sc&T c) → F cc ∀E, 3 5 ¬F cc ∀E, 2 6 ¬(Sc&T c) MT, 4, 5 7 ¬Sc ∨ ¬T c DeM, 6 8 Sc ∧E, 1 9 ¬T c ∨E, 7, 8 1 Laws of propositional logic: 2.1 – 2.3 Laws of Boolean algebra: Rules of inference known to be valid arguments: 5 QL Translations 10. Translate the following sentences into QL, providing a symbolization key and specifying the UD when necessary. (3 pts. each) (a) No Democrat likes every Republican. (b) Adams can’t do every job right. (c) Adams can’t do any job right. (d) Every farmer who owns a donkey needs hay. (e) There is no natural number less than zero. 6 QL Proofs 11. Construct a proof to show that the following argument is valid. (15 pts.) 3xFx + VyMy, Fg. Mg 7 Extra Credit 12. Translate the following argument into QL, providing a symbolization key and specifying the UD as necessary, then construct a proof to show that it’s valid. (10 pts.) The Mona Lisa is beautiful. It follows that anyone who steals the Mona Lisa steals something beautiful. A person deserves the death penalty if and only if he is a serial killer. Bundy is a serial killer, but Oswald is not. Both Bundy and Oswald are persons. Hence, Bundy deserves the death penalty, but Owald does not. X Х – X is 1 Px=x is a person Dx= deserves the death penalty Sx = x a senal killer b = Bundy 4x (PX (DxSx)) Oswald Sba-So QL Pb & Do Translation Proofs: Db &nDo 4+ (P+ (D x [5x)) 2 Sbar So :Db&rDo 3 LPb&Po 4 po → (Ob (56) HEI 5 pb 6bba Sh MP (JE) 4,5 7 Sb 8 Db 6,7 9 VEI som Elimation AE Modus Ponens XE 2 2 And/conjunction Elimination Biconditional Elimination 9 Pox (Dosso) 6,
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